# An investigation for working out hidden faces as different number of cubes are joined by making different shapes.

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Introduction

Amit Patel

Mathematics Coursework

An investigation for working out hidden faces as different number of cubes are joined by making different shapes.

As part of my GCSE mathematics requirements I have been assigned to investigate the number of hidden faces as cubes are joined in various way I shall start by making diagrams on the dotted paper provided and shall work out an expression (formula) which will reflect a relationship between the hidden faces and the number of cubes used.

Part 1: Number of cubes

- When I take a single cube and place it on dotted paper I can see 1 face hidden and faces visible from total of 6 faces.

- Similarly when I join 2 cubes and place them flat on a surface I can figure out that 4 faces are hidden and 8 are visible from a total of 12 cubes.

- When 3 cubes are joined to a similar pattern as mentioned above a total of 7 hidden faces and 11 visible faces out of a possible 18 faces observed

I am presenting a small table which describes the number of cubes used, the hidden faces, the visible faces and the total number of faces

nth term | No of cubes | No of hidden faces | No of visible faces | No of total faces |

1 | 1 | 1 | 5 | 6 |

2 | 2 | 4 | 8 | 12 |

3 | 3 | 7 | 11 | 18 |

4 | 4 | 10 | 14 | 24 |

5 | 5 | 13 | 17 | 30 |

6 | 6 | 16 | 20 | 36 |

7 | 7 | 19 | 23 | 42 |

From the above table a simple sequence can be formed and an nth term of the sequence can be worked out

Sequence for hidden faces

1 4 7 10 13 16 19

3 3 3 3 3 3

By taking the 1st

Middle

13 13 13 13 13 13

Sequence for visible faces

11 16 21 26 31 36 41 the nth term for the sequence is 5n+11

5 5 5 5 5 5

- When I join 5 cubes into 4 rows and lay it onto a flat surface I can find a total number of 120 faces with 38 of those showing and 82 hidden. The following table shows the sequence of 4 cubes.

nth term | No of cubes | No of hidden faces | No of visible faces | No of total faces |

1 | 4 | 10 | 14 | 24 |

2 | 8 | 28 | 20 | 48 |

3 | 12 | 46 | 26 | 72 |

4 | 16 | 64 | 32 | 96 |

5 | 20 | 82 | 38 | 120 |

6 | 24 | 100 | 44 | 144 |

7 | 28 | 118 | 50 | 168 |

From the table 2 simple sequences can be formed and an nth term of the sequence

Can be worked out

Sequence for hidden faces

10 28 46 64 82 100 118 the nth term for the sequence is 18n-8

18 18 18 18 18 18

Sequence for visible faces

14 20 26 31 38 44 50 the nth term for the sequence is 6n+8

6 6 6 6 6 6

- When five cubes are joined together in 5 rows I can observe 105 hidden faces, 45 showing faces with a total of 150 faces all together.

This table will show the next seven terms of the sequence

nth term | No of cubes | No of hidden faces | No of visible faces | No of total faces |

1 | 5 | 13 | 17 | 30 |

2 | 10 | 36 | 24 | 60 |

3 | 15 | 59 | 31 | 90 |

4 | 20 | 82 | 38 | 120 |

5 | 25 | 105 | 45 | 150 |

6 | 30 | 128 | 52 | 180 |

7 | 35 | 151 | 59 | 210 |

From the table 2 simple sequences can be formed and an nth term of the sequence

Can be worked out

Sequence for hidden faces

13 36 59 82 59 82 105 128 151 the nth term for the sequence is

23n-10

23 23 23 23 23 23 23 23

Sequence for visible faces

17 24 31 38 45 52 59 the nth term for the sequence is

7n+10

7 7 7 7 7 7

I have made this formula based on the above sequences which I have made by taking the height, width and the length of cuboids.

6 HWL-2HL-2HW-LW

Conclusion

No of total faces

1

9

27

27

54

2

18

72

36

108

3

27

117

45

162

4

36

162

54

216

5

45

207

63

270

6

54

252

72

324

7

63

297

81

378

From the above table a simple sequences can be formed and an nth term of the sequence

Can be worked out

Sequence for hidden faces

27 72 117 162 207 252 297 the nth term for the sequence 45n-18

45 45 45 45 45 45

Testifying the above sequences

Considering the formula used previously I am going to test the nth term made for bigger cuboids or cubes made from small cubes.

6 HWL-(LW+2HL+2HW)

1. If I take the nth term of a cuboid which I made previously as 7n-4 and test it for the fifth term then it is equal to 31. By using the general formula L = 5 H = 2 W = 1.

6 x 2 x 1 x 5 - (5 x 1 + 2 x 5 x 2 + 2 x 2 x 1)

60-29=31

2. By applying the nth term 18n-8 for the fifth term the answer comes to

18 x 5 - 8 = 82

By applying the final formula L = 5 H = 2 W = 2

6 x 5 x 2 x 2 - (5 x 2 + 2 x 2 x 5 + 2 x 2 x 2)

120 – 38 = 82

3. For my final example of a cuboid with the nth term of 45n-18 the value of the fifth term is 207 where as by applying the final formula for cuboids L = 5 H = 3 W = 3

6 x 5 x 3 x 3 - (5 x 3 + 2 x 3 x 5 + 2 x 3 x 3)

270 - 63 = 207

Evaluation

I conclude from various testing which I have conducted earlier that as the columns, rows and depth of a cuboid is altered the general formula produced will adjust itself accordingly and a true number of hidden faces can be worked out easily. This also explains the generalisations, predictions and other features considered during this assignment and strengthen the solution of hidden faces.

This student written piece of work is one of many that can be found in our GCSE Hidden Faces and Cubes section.

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